You have found the following ages (in years) of 6 bears. Those bears were randomly selected from the 29 bears at your local zoo: $ 42,\enspace 12,\enspace 24,\enspace 5,\enspace 22,\enspace 21$ Based on your sample, what is the average age of the bears? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we only have data for a small sample of the 29 bears, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{42 + 12 + 24 + 5 + 22 + 21}{{6}} = {21\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {441} + {81} + {9} + {256} + {1} + {0}} {{6 - 1}} $ {s^2} = \dfrac{{788}}{{5}} = {157.6\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{157.6\text{ years}^2}} = {12.6\text{ years}} $ We can estimate that the average bear at the zoo is 21 years old. There is also a standard deviation of 12.6 years.